What is the pOH of a 2.6 x 10-6 M H+ solution?

if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?

horsena [70]

Answer:

Silver Acetate would be the Limiting Reagent.

Explanation:

The balance chemical equation for the given double displacement reaction is as;

HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: Calculate Moles of Starting Materials:

Moles of HCl:

Moles = Mass / M.Mass

Moles = 72.9 g / 36.46

Moles = 1.99 moles

Moles of AgC₂H₃O₂:

Moles = 150 g / 166.91 g/mol

Moles = 0.898 moles

Step 2: Find out Limiting reagent as:

According to balance chemical equation.

1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂

So,

1.99 moles of HCl will react with = X moles of AgC₂H₃O₂

Solving for X,

X = 1.99 mol × 1 mol / 1 mol

X = 1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

8 0

2 years ago

Correct forms of the equation of Charles’s law is (are)

xxTIMURxx [149]

According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.

V∝ T, P is constant

where V, T and P are volume, temperature and pressure

$\frac{V1}{T1 }$ = $\frac{V2}{T2}$

where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.

8 0

2 years ago

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

8 0

2 years ago

What is the pressure in mm hg of a 0.025 mole sample of co2 at 350 k in a 200 l container?

Leni [432]

Answer:

0.65mmHg

Explanation:

Take R =8.314J/mol/K

V=200dm3

PV=nRT

P=0.025×8.314×623/200

=0.65mmHg

7 0

1 year ago

Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at

WITCHER [35]

Answer: The volume of given amount of ethanol at this temperature is 159.44 mL

Explanation:

Specific gravity is given by the formula:

$\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}$

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

$0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL$

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

$\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL$

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0

2 years ago